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3 years ago

Which is the correct name for the compound Ca(CLO2)2

2 answers:

6 0

The correct name is Calcium Chlorite. Calcium has a charge of 2 on the periodic table. In the equation, ClO has a subscript of 2, meaning that there is one less oxygen than chlor"ate", and so it is Chlor"ite". Hope this helped!

GalinKa [24]3 years ago

4 0

Calcium hypochlorite

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A sample was prepared by mixing 18. ml of 3.00 x 10^-3 m crystal violet (cv) with 2.00 ml of 0.250 m naoh. calculate the resulti

Aleks [24]

Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.

Explanation :

Step 1 : Find moles of crystal violet and NaOH.

The molarity formula is

$Molarity = \frac{mol}{L}$

Molarity of crystal violet = $3.00 \times 10^{-3} = \frac{mol (CrystalViolet)}{L}$

The volume of crystal violet solution is 18 mL which is 0.018 L.

Moles of crystal violet = $3.00 \times 10^{-3} \times 0.018 = 5.4 \times 10^{-5}$

Moles of crystal violet = 5.4 x 10⁻⁵

Moles of NaOH = $Molarity \times L = 0.250 \times 0.00200 = 5.00 \times 10^{-4}$

Moles of NaOH = 5.00 x 10⁻⁴

Step 2 : Find total volume of the solution

The total volume of the solution after mixing NaOH and crystal violet is

0.018 L + 0.00200 = 0.020 L

Step 3 : Use molarity formula to find final concentrations

Molarity of crystal violet = $\frac{mol(CrystalViolet)}{Total Volume(L) } = \frac{5.4 \times 10^{-5}}{0.020} = 2.7 \times 10^{-3}$

Final concentration of CV = 0.0027 M

Molarity of NaOH= $\frac{mol(NaOH)}{Total Volume(L) } = \frac{5.00 \times 10^{-4}}{0.020} = 0.025 \times 10^{-3}$

NaOH is a strong base and dissociates completely as follows.

$NaOH (aq) \rightarrow Na^{+} (aq) + OH^{-} (aq)$

The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.

Concentration of OH⁻ = 0.025 M

8 0

3 years ago

What is the correct answer?!????

ddd [48]

Answer:

Explanation:

Dalton worked with mainly about the chemistry of atoms.

how do atoms combine to form various molecules.

—rather than the details of the physical, internal structure of atoms, although he never denied the possibility of atoms' having a substructure.

4 0

4 years ago

Read the chemical equation. Fe2O3 + CO → Fe + CO2 If 2 moles of Fe2O3 react with 9 moles of CO, how many moles of each product a

cluponka [151]

Answer:

4 moles Fe and 6 moles CO2 are the moles of each products formed

Explanation:

Fe2O3 + 3CO → 2Fe + 3CO2

1st step Ballance the equation. Afterwards, you can work properly

1 mol of Fe2O3 reacts with 3 moles of CO to make 2 moles of Fe and 3 moles of CO2

2nd step Predict the reactant in excess and limitant reagent.

If 1 moles of Fe2O3 reacts with 3 moles of CO

2 moles of Fe2O3 reacts with 6 moles of CO (2.3) /1

I have 9 moles of CO, so the Fe2O3 is my limitant reagent.

<u><em>REMEMBER</em></u> you always have to work with the limitant.

If 3 moles of CO reacts with 1 mol of Fe2O3

9 moles of CO reacts with 3 moles of Fe2O3 (9.1) /3

I have 2 moles of Fe2O3, so I still have Fe2O3, by the way the CO is the reactant in excess. (Just to show all)

3rd step Work with the limitant reagent.

1 mol of Fe2O3 ___ makes____ 2 moles of Fe + 3 moles of CO

2 mol of Fe2O3 ___ makes ___ 4 moles of Fe + 6 moles of CO

5 0

3 years ago

Coal containing 15.0% H2O, 2.0% S and 83.0% C by mass is burnt with the stoichiometric amount of air in a furnace. What is the m

devlian [24]

Answer:

This is a coal combustion process and we will assume

Inlet coal amount = 100kg

It means that there are

15kg of H2O, 2kg of Sulphur and 83kg of Carbon

Now to find the mole fraction of SO2(g) in the exhaust?

Molar mass of S = 32kg/kmol

Initial moles n of S = 2/32 = 0.0625kmols

Reaction: S + O₂ = SO₂

That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂

Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust

The mole fraction of SO2(g) in the exhaust=0.0625kmols

Explanation:

5 0

3 years ago

In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) ch

LekaFEV [45]

Answer : The volume of stock solution needed are, 12.5 mL

Explanation :

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of copper (II) chloride.

$M_2\text{ and }V_2$ are the final molarity and volume of stock solution of copper (II) chloride.

We are given:

$M_1=0.100M\\V_1=250.0mL\\M_2=2.00M\\V_2=?$

Putting values in above equation, we get:

$0.100M\times 250.0mL=2.00M\times V_2\\\\V_2=12.5mL$

Hence, the volume of stock solution needed are, 12.5 mL

8 0

3 years ago